Answer:
Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing
Explanation:
The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item
Answer:
The Full details of the answer is attached.
Answer: they would be 300 miles from the station.
Explanation:
At the point where both trains meet, they would have covered the same distance.
Let t represent the time spent by the first train in covering this distance.
Distance = speed × time
The first train leaves the station and travels north at 60km/hr.
Distance covered by the first train is
60 × t = 60t
Two hours later, a second train leaves on a parallel track and travels north at 100km/hr. Time spent by the second train in covering this distance is (t - 2) hours
Distance covered by the second train is
100(t - 2) = 100t - 200
Since both trains covered the same distance, then
100t - 200 = 60t
100t - 60t = 200
40t = 200
t = 200/40
t = 5 hours
The distance that they would be from the station is
60 × 5 = 300 miles
Answer:
The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.
Explanation:
The net force over the window is calculated by multiplying the difference in pressure by the area of the window:
F = Δp*A
The pressure inside the plane is around 1 atm, hence the difference in pressure is:
Δp = 1atm - 0.35 atm = 0.65 atm
Expressing in the EE unit system:
Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2
Replacing in the force:
F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf
For the international unit system, we re-calculate the window's area and the difference in pressure:
A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2
Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2
Replacing in the force:
F = 65861 N/m^2 *0.0516 m^2 = 3398 N
